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%% 填写课程信息：
\newcommand{\CourseName}{高等代数复习2}
\newcommand{\CourseStudents}{2024 级数学与应用数学1班}
\newcommand{\ExamContents}{本次数学考试的主要内容是：
}

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\begin{document}

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\date{2024年12月26日} %考试日期

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\author{学号 \underline{\hspace{3.5cm}} 姓名 \underline{\hspace{3.5cm}} }
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\author{\CourseStudents}
\title{\CourseName 解答 }
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%选择题开始
\begin{enumerate}

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%\newpage 
\item  %选择题第1题
齐次线性方程组 $AX=0$ 有非零解的充要条件是哪个？
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%
\begin{tasks}(1) %每行1个选项
\task [A.]  $A$ 的任两个列向量线性相关。
\task [B.]  $A$ 中必有一个列向量是其余列向量的线性组合。
\task [C.]  $A$ 的任两个列向量线性无关。
\task [D.]  $A$ 中任意一个列向量都是其余向量的线性组合。
\end{tasks}
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{\color{red} 解答：B. 
将矩阵 $A$ 按照列向量的方式写成 $A=(\alpha_1, \alpha_2,\cdots,\alpha_n)$,  则齐次线性方程组 $AX=0$ 可以写成
$x_1\alpha_1 + x_2\alpha_2 + \cdots + x_n\alpha_n =0. $
因此齐次线性方程组 $AX=0$ 有非零解等价于矩阵 $A$ 的列向量组线性相关。 因此矩阵 $A$ 的某个列向量是其余列向量的线性组合。

}

\fi

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%\newpage 
\item  %选择题第2题
关于非齐次线性方程组 $AX=\beta$ 和其导出组 $AX=0$, 下列命题成立的是哪个？
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%
\begin{tasks}(1) %每行1个选项
\task [A.]  $AX=0$ 只有零解时，$AX=\beta$ 有唯一解。
\task [B.]  $AX=0$ 有非零解时，$AX=\beta$ 有无穷多解。
\task [C.]  $AX=\beta$ 有唯一解时，$AX=0$ 只有零解。
\task [D.]  $AX=\beta$ 无解时，$AX=0$ 也无解。
\end{tasks}
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{\color{red} 解答：C. 设 $A$ 是 $m\times n$ 矩阵。此时 $X=(x_1,x_2,\cdots, x_n)^T$. 
\begin{tasks}(1) %每行2个选项
\task [A.]  若 $AX=0$ 只有零解，则 $R(A)=n$. 若 $m>n$, 则有可能增广矩阵 $(A,\beta)$ 的秩大于系数矩阵 $A$ 的秩，这时 $AX=\beta$ 无解。若仍有 $R(A,\beta)=R(A)=n$, 则 $AX=\beta$ 有唯一解。
\task [B.]  当 $AX=0$ 有非零解时，$R(A)<n$. 这时 $R(A,\beta)=R(A)$ 与 $R(A,\beta)=R(A)+1$ 都有可能，因此 $AX=\beta$ 有无穷多解或无解。
\task [C.]  当 $AX=\beta$ 有唯一解时，$R(A,\beta)=R(A)=n$. 这时 $AX=0$ 只有零解。
\task [D.]  齐次线性方程组 $AX=0$ 总是有解的，因为 $X=0$ 总是 $AX=0$ 的解。
\end{tasks}
}

\fi

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%\newpage 
\item  %选择题第3题
设 $n$ 阶方阵 $A$ 满足 $A^2-2A=O$，则下列不一定可逆的矩阵是哪个？
%
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\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(4) %每行4个选项
\task [A.]  $A-3E_n$ 
\task [B.]  $A-E_n$ 
\task [C.]  $A+2E_n$
\task [D.]  $A$
\end{tasks}
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{\color{red} 解答：D. 
\begin{tasks}(1) %每行1个选项
\task [A.]  因为 $(A-3E)(A+E)=-3E$, 所以 $A-3E$ 可逆。
\task [B.]  因为 $(A-E)(A-E)=E$,  所以 $A-E$ 可逆。
\task [C.]  因为 $(A+2E)(A-4E)=-8E$, 所以 $A+2E$ 可逆。
\task [D.]  题目条件是 $A(A-2E)=O$. 当$A=2E$ 时满足题目条件，这时 $A$可逆；但当 $A=O$ 时也满足题目条件，这时 $A$ 不可逆。
\end{tasks}

}

\fi

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%\newpage 
\item  %选择题第4题
已知 4 元非齐次线性方程组 $AX=\beta$，$R(A)=3$，$\eta_1, \eta_2, \eta_3$ 是它的三个解向量，其中\\ $\eta_1+\eta_2=(1,2,0,2)^T, \eta_2+\eta_3=(1,0,1,3)^T$，求其导出组的通解。
%
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\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(4) %每行4个选项
\task [A.]  $k(2,2,1,5)^T$. 
\task [B.]  $k(1,2,0,2)^T$. 
\task [C.]  $k(1,0,1,3)^T$. 
\task [D.]  $k(0,2,-1,-1)^T$. 
\end{tasks}

\vspace{0.2cm}

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{\color{red} 解答：D. 
根据 $n-R(A)=4-3=1$ 可知导出组 $AX=0$ 的基础解系包含1个向量。
因为 $A\eta_1=\beta, A\eta_3=\beta$, 所以 $A(\eta_1-\eta_3)=0$, 所以 $\eta_1-\eta_3$ 是 $AX=0$ 的基础解系。
$\eta_1-\eta_3 = (\eta_1+\eta_2)-(\eta_2+\eta_3)=(1,2,0,2)^T-(1,0,1,3)^T=(0,2,-1,-1)^T$. 

 }

\fi

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%\newpage 
\item  %选择题第5题
向量组 $\alpha_1=(1,2,3)^T, \alpha_2=(3,-1,2)^T, \alpha_3=(2,3,x)^T$ 线性无关，求 $x$ 的值。 
%
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\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(4) %每行4个选项
\task [A.]  $x=5$. 
\task [B.]  $x\neq 5$. 
\task [C.]  $x=0$.
\task [D.]  $x\neq 0$. 
\end{tasks}

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{\color{red} 解答：B. 
设向量组 $\alpha_1, \alpha_2, \alpha_3$ 线性无关，则根据定义，向量方程 $k_1\alpha_1+k_2\alpha_2+k_3\alpha_3=0$ 只有零解 $k_1=k_2=k_3=0$. 此即线性方程组
$$
\begin{pmatrix} 1&3&2 \\ 2&-1&3 \\ 3&2&x  \end{pmatrix} 
\begin{pmatrix}k_1 \\ k_2 \\ k_3 \end{pmatrix} 
=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} 
$$
只有零解。因此系数行列式不等于零。因此 $x\neq 5$. 

}


\fi

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%\newpage 
\item  %选择题第6题
向量组 $\alpha_1=(1,2,-1,1)^T, \alpha_2=(2,0,t,0)^T, \alpha_3=(0,4,-5,2)^T$ 的秩为 2，求 $t$ 的值。 
%
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\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(4) %每行4个选项
\task [A.]  $1$. 
\task [B.]  $2$.
\task [C.]  $3$.
\task [D.]  $4$. 
\end{tasks}

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\ifnum\showsolution>0

{\color{red} 解答：C. 
将向量 $\alpha_1, \alpha_3, \alpha_2$ 按行向量排列成矩阵，（注意到将含参数的向量放在后面，可以简化处等行变换），然后用初等行变换化为行阶梯形，可得
\begin{equation*}
\begin{pmatrix} 1&2&-1&1 \\ 0&4&-5&2 \\ 2&0&t&0 \end{pmatrix}
\xrightarrow[]{}
\begin{pmatrix} 1&2&-1&1 \\ 0&4&-5&2 \\ 0&-4&t+2&-2 \end{pmatrix}
\xrightarrow[]{}
\begin{pmatrix} 1&2&-1&1 \\ 0&4&-5&2 \\ 0&0&t-3&0 \end{pmatrix}.
\end{equation*}
可见当 $t=3$ 时，向量组的秩为2. 

}


\fi

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%\newpage 
\item  %选择题第7题
已知向量组 $\alpha_1=(1,0,1)^T, \alpha_2=(2,2,3)^T, \alpha_3=(2,6,t)^T$ 线性相关，求 $t$ 的值。
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(4) %每行4个选项
\task [A.]  $3$. 
\task [B.]  $4$. 
\task [C.]  $5$. 
\task [D.]  $6$. 
\end{tasks}

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\ifnum\showsolution>0

{\color{red} 解答：C. 
因为这三个向量线性相关，所以其秩小于3. 使用初等行变换将这些向量排列成的矩阵化成行阶梯形，可得
\begin{equation*}
\begin{pmatrix} 1&0&1 \\ 2&2&3 \\ 2&6&t \end{pmatrix}
\xrightarrow[]{}
\begin{pmatrix} 1&0&1 \\ 0&2&1 \\ 0&6&t-2 \end{pmatrix}
\xrightarrow[]{}
\begin{pmatrix} 1&0&1 \\ 0&2&1 \\ 0&0&t-5 \end{pmatrix}. 
\end{equation*}
可见当 $t=5$ 时这个矩阵的秩小于3. （另解：令这三个向量排列成的行列式的值等于零，也可求得 $t=5$. 上述初等行变换化为阶梯形的方法与求行列式的值的方法本质是一样的。）

}

\fi

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%\newpage 
\item  %选择题第8题
设有如下线性方程组，下述说法中不正确的是哪个？
%
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\dotfill (\hspace{1cm})
\fi
%
\begin{equation*}
\left\{
\begin{aligned}
(2-k)x_1+2x_2-2x_3 &=1\\
2x_1+(5-k)x_2-4x_3 &=2 \\
-2x_1-4x_2+(5-k)x_3 &=-k-1 \\
\end{aligned}
\right. 
\end{equation*}

\begin{tasks}(1) %每行1个选项
\task [A.]  当 $k=10$ 时，线性方程组无解。
\task [B.]  当 $k=1$ 时，线性方程组有无穷多解。
\task [C.]  当 $k\neq 1,10$ 时，线性方程组有唯一解。
\task [D.]  以上说法都不正确。 
\end{tasks}

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\ifnum\showsolution>0

{\color{red} 解答：D. 
将增广矩阵用初等行变换化为阶梯形矩阵，并用线性方程组有解的判别定理，求解本题。（注意将未知参数放在增广矩阵的右下角，可以方便计算。）
\begin{equation*}
\begin{aligned}
& \begin{pmatrix} -2&-4&5-k&-k-1 \\ 2&5-k&-4&2 \\ 2-k&2&-2&1 \end{pmatrix}
\xrightarrow[]{}
\begin{pmatrix} -2&-4&5-k&-k-1 \\ 0&1-k&1-k&1-k \\ 4-2k&4&-4&2 \end{pmatrix}\\ 
\xrightarrow[]{}
&\begin{pmatrix} -2&-4&5-k&-k-1 \\ 0&1-k&1-k&1-k \\ 0&-4+4k&6-7k+k^2&k^2-k \end{pmatrix}
\xrightarrow[]{}
\begin{pmatrix} -2&-4&5-k&-k-1 \\ 0&1-k&1-k&1-k \\ 0&0&k^2-11k+10&k^2-5k+4 \end{pmatrix} \\ 
=&\begin{pmatrix} -2&-4&5-k&-k-1 \\ 0&1-k&1-k&1-k \\ 0&0&(k-1)(k-10)&(k-1)(k-4) \end{pmatrix}. 
\end{aligned}
\end{equation*}
因此当 $k=1$ 时，$R(\overline{A})=R(A)=1$, 而未知数个数 $n=3$, 所以线性方程组有无穷多解。\\ 
当 $k=10$ 时，$R(A)=2, R(\overline{A})=3$, 所以线性方程组无解。\\ 
当 $k\neq 1$ 且 $k\neq 10$ 时，$R(\overline{A})=R(A)=3$, 所以线性方程组有唯一解。
}

\fi

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%\newpage 
\item  %选择题第9题
设 $A$ 为 $4$ 阶方阵，且 $|A|=2$, 则 $|2AA^*|$ 的值等于多少？
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\dotfill (\hspace{1cm})
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%
\begin{tasks}(4) %每行4个选项
\task [A.]  $64$. 
\task [B.]  $256$. 
\task [C.]  $512$. 
\task [D.]  $1024$. 
\end{tasks}

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%%选择题第9题：解答与评分标准
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{\color{red} 解答：B.
根据伴随矩阵的性质，可得
$$2AA^* = 2|A|E_4=4E_4= \begin{pmatrix} 4&0&0&0 \\ 0&4&0&0 \\ 0&0&4&0 \\ 0&0&0&4 \end{pmatrix}. $$
因此其行列式的值为 $4^4=256$. 

}

\fi

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%\newpage 
\item  %选择题第10题
设矩阵 $A$ 如下，则矩阵 $(A^*)^{-1}$ 等于什么？
%
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\dotfill (\hspace{1cm})
\fi
%
$$A=\begin{pmatrix}
1 & 2 & -1 \\
3 & 4 & -2 \\
5 & 3 & 1 \\
\end{pmatrix}. 
$$
\begin{tasks}(4) %每行4个选项
\task [A.]  $(A^*)^{-1}=-\frac{1}{5}A$. 
\task [B.]  $(A^*)^{-1}=\frac{1}{5}A$.  
\task [C.]  $(A^*)^{-1}=-5A$.  
\task [D.]  $(A^*)^{-1}=5A$.  
\end{tasks}

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%%选择题第10题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：A. 
设 $d=|A|$, 根据 $AA^*=dE$ 可得 $A^*=dA^{-1}$ 以及 $(A^*)^{-1} = \frac{1}{d}A$. 
计算可得 $d=-5$. 

}


\fi
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%\newpage 
\item  %选择题第11题
设 $ABA^{-1}=BA^{-1}+3E$, 则矩阵 $B$ 为
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(4) %每行4个选项
\task [A.]  $3(A-E)^{-1}A$. 
\task [B.]  $(A-3E)^{-1}A$. 
\task [C.]  $3(A-E)A^{-1}$. 
\task [D.]  $(A-3E)A^{-1}$. 
\end{tasks}

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%%选择题第11题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：A. 
\begin{equation*}
\begin{aligned}
ABA^{-1} = BA^{-1} + 3E 
\Rightarrow  
(A-E)BA^{-1} = 3E  
\Rightarrow  
(A-E)B = 3A 
\Rightarrow  
B = 3(A-E)^{-1}A.  
\end{aligned}
\end{equation*}

}


\fi

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%\newpage 
\item  %选择题第12题

设矩阵 $A$ 的伴随矩阵 $A^*$ 如下，则矩阵 $A$ 的所有元素的和等于多少？
%
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%
$$ 
A^*=\begin{pmatrix}
1 & 0 &0 & 0\\
0 & 1 & 0 & 0\\
1 & 0 & 1 & 0\\
0 & -3 & 0 & 8\\
\end{pmatrix}. 
$$

\begin{tasks}(4) %每行4个选项
\task [A.]  $5$. 
\task [B.]  $6$.
\task [C.]  $7$.
\task [D.]  $8$. 
\end{tasks}

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%%选择题第12题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：A. 
设 $d=|A|$, 根据 $AA^*=dE_4$ 可得 $|AA^*|=d^4$. 所以 $|A^*|=d^3$. 计算可得 $|A^*|=8$, 所以 $d=2$. 
根据 $AA^*=dE_4$ 又可得 $A=d(A^*)^{-1}=2(A^*)^{-1}$. 因此问题化为计算 $A^*$ 的逆阵。 
将 $(A^*,E_4)$ 通过初等行变换化为 $(E_4,(A^*)^{-1})$，可得 
$$(A^*)^{-1} = \begin{pmatrix}
1 & 0 &0 & 0\\
0 & 1 & 0 & 0\\
-1 & 0 & 1 & 0\\
0 & \frac{3}{8} & 0 & \frac{1}{8} \\
\end{pmatrix}.
$$
因此矩阵 $A$ 的所有元素的和为 $2(1+1+1-1+\frac{3}{8}+\frac{1}{8})=5$. 

}

\fi

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%选择题结束
\end{enumerate}


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\end{document}





